Transcript Slide 1

3
Systems of Linear
Equations and Matrices
Copyright © Cengage Learning. All rights reserved.
3.2
Using Matrices to Solve Systems of Equations
Copyright © Cengage Learning. All rights reserved.
Using Matrices to Solve Systems of Equations
In this section we describe a systematic method for solving
systems of equations that makes solving large systems of
equations in any number of unknowns straightforward.
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Using Matrices to Solve Systems of Equations
First, some terminology:
Linear Equation
A linear equation in the n variables x1, x2, . . . , xn has the
form
a1x1 + . . . + anxn = b. (a1, a2, . . . , an, b constants)
The numbers a1, a2, . . . , an are called the coefficients,
and the number b is called the constant term, or
right-hand side.
Quick Example
1. 3x – 5y = 0
Linear equation in x and y
Coefficients: 3, −5 Constant term: 0
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Using Matrices to Solve Systems of Equations
Notice that a linear equation in any number of unknowns
(for example, 2x – y = 3) is entirely determined by its
coefficients and its constant term.
In other words, if we were simply given the row of numbers
[2 –1 3]
we could easily reconstruct the original linear equation by
multiplying the first number by x, the second by y, and
inserting a plus sign and an equals sign, as follows:
2  x + (–1)  y = 3
or
2x – y = 3.
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Using Matrices to Solve Systems of Equations
Similarly, the equation
–4x + 2y = 0
is represented by the row
[–4 2 0],
and the equation
is represented by
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Using Matrices to Solve Systems of Equations
As the last example shows, the first number is always the
coefficient of x and the second is the coefficient of y. If an x
or a y is missing, we write a zero for its coefficient.
We shall call such a row the coefficient row of an
equation. If we have a system of equations, for example
the system
2x – y = 3
–x + 2y = –4,
we can put the coefficient rows together like this:
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Using Matrices to Solve Systems of Equations
We call this the augmented matrix of the system of
equations. The term “augmented” means that we have
included the right-hand sides 3 and –4.
We will often drop the word “augmented” and simply refer
to the matrix of the system.
A matrix (plural: matrices) is nothing more than a
rectangular array of numbers as above.
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Using Matrices to Solve Systems of Equations
Matrix, Augmented Matrix
A matrix is a rectangular array of numbers.The
augmented matrix of a system of linear equations is the
matrix whose rows are the coefficient rows of the
equations.
Quick Example
The augmented matrix of the system
x+y=3
x–y=1
is
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Using Matrices to Solve Systems of Equations
Here is the same operation both in the language of
equations and the language of rows. (We refer to the
equation here as Equation 1, or simply E1 for short, and to
the row as Row 1, or R1.)
Multiplying both sides of an equation by the number a
corresponds to multiplying the coefficient row by a.
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Using Matrices to Solve Systems of Equations
Now look at what we do when we add two equations:
All we are really doing is adding the corresponding entries
in the rows, or adding the rows.
In other words,
Adding two equations corresponds to adding their
coefficient rows.
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Using Matrices to Solve Systems of Equations
In short, the manipulations of equations that we have seen
can be done more easily with rows in a matrix because we
don’t have to carry x, y, and other unnecessary notation
along with us; x and y can always be inserted at the end if
desired.
The manipulations we are talking about are known as row
operations.
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Using Matrices to Solve Systems of Equations
In particular, we use three elementary row operations.
Elementary Row Operations
Type 1: Replacing Ri by aRi (where a  0)
In words: multiplying or dividing a row by a nonzero number.
Type 2: Replacing Ri by aRi  bRj (where a  0)
Multiplying a row by a nonzero number and adding or
subtracting a multiple of another row.
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Using Matrices to Solve Systems of Equations
Type 3: Switching the order of the rows
This corresponds to switching the order in which we write
the equations; occasionally this will be convenient.
For Types 1 and 2, we write the instruction for the row
operation next to the row we wish to replace.
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Using Matrices to Solve Systems of Equations
Quick Examples
Replace R2 by 3R2.
Replace R1 by
4R1 – 3R2.
Switch R1 and R2.
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Using Matrices to Solve Systems of Equations
One very important fact about the elementary row
operations is that they do not change the solutions of the
corresponding system of equations.
In other words, the new system of equations that we get by
applying any one of these operations will have exactly the
same solutions as the original system: It is easy to see that
numbers that make the original equations true will also
make the new equations true, because each of the
elementary row operations corresponds to a valid operation
on the original equations.
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Using Matrices to Solve Systems of Equations
That any solution of the new system is a solution of the old
system follows from the fact that these row operations are
invertible: The effects of a row operation can be reversed
by applying another row operation, called its inverse.
Here are some examples of this invertibility.
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Using Matrices to Solve Systems of Equations
Our objective, then, is to use row operations to change the
system we are given into one with exactly the same set of
solutions, in which it is easy to see what the solutions are.
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Solving Systems of Equations by
Using Row Operations
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Solving Systems of Equations by Using Row Operations
Now we put rows to work for us in solving systems of
equations. Let’s start with a complicated looking system of
equations:
We begin by writing the matrix of the system:
Now what do we do with this matrix?
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Solving Systems of Equations by Using Row Operations
Step 1 Clear the fractions and/or decimals (if any)
using operations of Type 1.
To clear the fractions, we multiply the first row by 6 and the
second row by 4.
We record the operations by writing the symbolic form of an
operation next to the row it will change, as follows.
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Solving Systems of Equations by Using Row Operations
By this we mean that we will replace the first row by 6R1
and the second by 4R2. Doing these operations gives
Step 2 Designate the first nonzero entry in the first row
as the pivot.
In this case we designate the entry –4 in the first row as the
“pivot” by putting a box around it:
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Solving Systems of Equations by Using Row Operations
Step 3 Use the pivot to clear its column using
operations of Type 2.
By clearing a column, we mean changing the matrix so
that the pivot is the only nonzero number in its column.
The procedure of clearing a column using a designated
pivot is also called pivoting.
Desired row 2 (the “#”s stand for as yet
unknown numbers)
Cleared pivot column
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Solving Systems of Equations by Using Row Operations
We want to replace R2 by a row of the form aR2  bR1 to
get a zero in column 1.
Moreover—and this will be important when we discuss the
simplex method we are going to choose positive values for
both a and b.
We need to choose a and b so that we get the desired
cancellation. We can do this quite mechanically as follows:
a. Write the name of the row you need to change on the left
and that of the pivot row on the right.
R2
R1
Row to change
Pivot row
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Solving Systems of Equations by Using Row Operations
b. Focus on the pivot column,
Multiply each row by
the absolute value of the entry currently in the other. (We
are not permitting a or b to be negative.)
4R2
From Row 1
1R1
From Row 2
The effect is to make the two entries in the pivot column
numerically the same. Sometimes, you can accomplish this
by using smaller values of a and b.
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Solving Systems of Equations by Using Row Operations
c. If the entries in the pivot column have opposite signs,
insert a plus (+). If they have the same sign, insert a
minus (–). Here, we get the instruction
4R2 + 1R1, or simply 4R2 + R1
d. Write the operation next to the row you want to change,
and then replace that row using the operation:
We have cleared the pivot column and completed Step 3.
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Solving Systems of Equations by Using Row Operations
Simplification Step (Optional) If, at any stage of the
process, all the numbers in a row are multiples of an
integer, divide by that integer—a Type 1 operation.
This is an optional but extremely helpful step: It makes the
numbers smaller and easier to work with.
In our case, the entries in R2 are divisible by 13, so we
divide that row by 13. (Alternatively, we could divide by
–13. Try it.)
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Solving Systems of Equations by Using Row Operations
Step 4 Select the first nonzero number in the second
row as the pivot, and clear its column.
Here we have combined two steps in one: selecting the
new pivot and clearing the column (pivoting). The pivot is
shown below, as well as the desired result when the
column has been cleared:
We now wish to get a 0 in place of the 3 in the pivot
column.
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Solving Systems of Equations by Using Row Operations
Let’s run once again through the mechanical steps to get
the row operation that accomplishes this.
a. Write the name of the row you need to change on the left
and that of the pivot row on the right:
R1
R2
Row to change
Pivot row
b. Focus on the pivot column,
Multiply each row by
the absolute value of the entry currently in the other:
1R1
3R2
From Row 2
From Row 1
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Solving Systems of Equations by Using Row Operations
c. If the entries in the pivot column have opposite signs,
insert a plus (+). If they have the same sign, insert a
minus (–). Here, we get the instruction
1R1 + 3R2.
d. Write the operation next to the row you want to change
and then replace that row using the operation.
Now we are essentially done, except for one last step.
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Solving Systems of Equations by Using Row Operations
Final Step Using operations of Type 1, turn each pivot
(the first nonzero entry in each row) into a 1.
We can accomplish this by dividing the first row by –4 and
multiplying the second row by –1:
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Solving Systems of Equations by Using Row Operations
The matrix now has the following nice form:
(This is the form we will always obtain with two equations in
two unknowns when there is a unique solution.) This form
is nice because, when we translate back into equations, we
get
1x + 0y = 3
0x + 1y = –2.
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Solving Systems of Equations by Using Row Operations
In other words,
x = 3 and y = –2
and so we have found the solution, which we can also write
as (x, y) = (3, –2).
The procedure we’ve just demonstrated is called
Gauss-Jordan reduction or row reduction.
It may seem too complicated a way to solve a system of
two equations in two unknowns, and it is. However, for
systems with more equations and more unknowns, it is
very efficient.
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Solving Systems of Equations by Using Row Operations
In Example 1 below we use row reduction to solve a
system of linear equations in three unknowns: x, y, and z.
Just as for a system in two unknowns, a solution of a
system in any number of unknowns consists of values for
each of the variables that, when substituted, satisfy all of
the equations in the system.
Again, just as for a system in two unknowns, any system of
linear equations in any number of unknowns has either no
solution, exactly one solution, or infinitely many solutions.
There are no other possibilities.
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Solving Systems of Equations by Using Row Operations
Solving a system in three unknowns graphically would
require the graphing of planes (flat surfaces) in three
dimensions. (The graph of a linear equation in three
unknowns is a flat surface.)
The use of row reduction makes three-dimensional
graphing unnecessary.
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Example 1 – Solving a System by Gauss-Jordan Reduction
Solve the system
x – y + 5z = –6
3x + 3y – z = 10
x + 3y + 2z = 5.
Solution:
The augmented matrix for this system is
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Example 1 – Solution
cont’d
Note that the columns correspond to x, y, z, and the righthand side, respectively.
We begin by selecting the pivot in the first row and clearing
its column.
Remember that clearing the column means that we turn all
other numbers in the column into zeros.
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Example 1 – Solution
cont’d
Thus, to clear the column of the first pivot, we need to
change two rows, setting up the row operations in exactly
the same way as above.
Notice that both row operations have the required form
aRc  bR1
Row to change
Pivot row
with a and b both positive.
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Example 1 – Solution
cont’d
Now we use the optional simplification step to simplify R2:
Next, we select the pivot in the second row and clear its
column:
R1 and R3 are to be changed.
R2 is the pivot row.
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Example 1 – Solution
cont’d
We simplify R3.
Now we select the pivot in the third row and clear its
column:
R1 and R2 are to be changed.
R3 is the pivot row.
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Example 1 – Solution
cont’d
Finally, we turn all the pivots into 1s:
The matrix is now reduced to a simple form, so we
translate back into equations to obtain the solution:
x = 1, y = 2, z = –1, or (x, y, z) = (1, 2, –1).
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Solving Systems of Equations by Using Row Operations
Notice the form of the very last matrix in the example:
The 1s are on the (main) diagonal of the matrix; the goal
in Gauss-Jordan reduction is to reduce our matrix to this
form.
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Solving Systems of Equations by Using Row Operations
Reduced Row Echelon Form
A matrix is said to be in reduced row echelon form or to
be row-reduced if it satisfies the following properties.
P1. The first nonzero entry in each row (called the leading
entry of that row) is a 1.
P2. The columns of the leading entries are clear (i.e., they
contain zeros in all positions other than that of the
leading entry).
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Solving Systems of Equations by Using Row Operations
P3. The leading entry in each row is to the right of the
leading entry in the row above, and any rows of zeros
are at the bottom.
Quick Examples
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The Traditional Gauss-Jordan
Method
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The Traditional Gauss-Jordan Method
In the version of the Gauss-Jordan method we have
presented, we eliminated fractions and decimals in the first
step and then worked with integer matrices, partly to make
hand computation easier and partly for mathematical
elegance.
However, complicated fractions and decimals present no
difficulty when we use technology.
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The Traditional Gauss-Jordan Method
The following example illustrates the more traditional
approach to Gauss-Jordan reduction used in many of the
computer programs that solve the huge systems of
equations that arise in practice.
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Example 7 – Solving a System with the Traditional Gauss-Jordan Method
Solve the following system using the traditional
Gauss-Jordan method:
2x + y + 3z = 5
3x + 2y + 4z = 7
2x + y + 5z = 10.
Solution:
We make two changes in our method. First, there is no
need to get rid of decimals.
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Example 7 – Solution
cont’d
Second, after selecting a pivot, divide the pivot row by the
pivot value, turning the pivot into a 1.
It is easier to determine the row operations that will clear
the pivot column if the pivot is a 1.
If we use technology to solve this system of equations, the
sequence of matrices might look like this:
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Example 7 – Solution
The solution is (x, y, z) = (–2, 1.5, 2.5).
cont’d
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